A bag contains 18 balls out of which x balls are white.

(i) If one ball is drawn at random from the bag, what is the probability that it is white ball?

(ii) If 2 more white balls are put in the bag, the probability of drawing a white ball will be $\dfrac{9}{8}$ times that of probability of white ball coming in part (i). Find the value of x.

(i) Let A be the event 'a white ball is drawn'.

∴ The no. of favourable outcomes to event A = x.

∴ P(a white ball is drawn) = $\dfrac{x}{18}.$

Hence, the probability that a white ball is drawn = $\dfrac{x}{18}$.

(ii) If 2 more white balls are added,

No. of white balls = x + 2,

Total no. of balls = 18 + 2 = 20.

Hence now,

P(a white ball is drawn) = $\dfrac{x + 2}{20}$

Given, probability of drawing a white ball will be $\dfrac{9}{8}$ times that of probability of white ball coming in part (i)

$\therefore \dfrac{x + 2}{20} = \dfrac{9}{8} \times \dfrac{x}{18} \\[1em] \Rightarrow \dfrac{x + 2}{20} = \dfrac{x}{16} \\[1em] \Rightarrow 16(x + 2) = 20x \\[1em] \Rightarrow 16x + 32 = 20x \\[1em] \Rightarrow 20x - 16x = 32 \\[1em] \Rightarrow 4x = 32 \\[1em] \Rightarrow x = 8.$

Hence, the value of x = 8.